$ B = \left[\begin{array}{rrr}-2 & 5 & 2 \\ 4 & -1 & 4\end{array}\right]$ $ C = \left[\begin{array}{rr}0 & -2 \\ 2 & 5 \\ 3 & 4\end{array}\right]$ What is $ B C$ ?
Answer: Because $ B$ has dimensions $(2\times3)$ and $ C$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B C = \left[\begin{array}{rrr}{-2} & {5} & {2} \\ {4} & {-1} & {4}\end{array}\right] \left[\begin{array}{rr}{0} & \color{#DF0030}{-2} \\ {2} & \color{#DF0030}{5} \\ {3} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{0}+{5}\cdot{2}+{2}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{0}+{5}\cdot{2}+{2}\cdot{3} & ? \\ {4}\cdot{0}+{-1}\cdot{2}+{4}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{0}+{5}\cdot{2}+{2}\cdot{3} & {-2}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{4} \\ {4}\cdot{0}+{-1}\cdot{2}+{4}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{0}+{5}\cdot{2}+{2}\cdot{3} & {-2}\cdot\color{#DF0030}{-2}+{5}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{4} \\ {4}\cdot{0}+{-1}\cdot{2}+{4}\cdot{3} & {4}\cdot\color{#DF0030}{-2}+{-1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}16 & 37 \\ 10 & 3\end{array}\right] $